Answer:
Approximately [tex]0.64\; {\rm s}[/tex].
Velocity: approximately [tex]6.3\; {\rm m\cdot s^{-1}}[/tex].
(Assuming that air resistance is negligible, and that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] assuming that the fall started from rest.)
Explanation:
If air resistance is negligible, acceleration will be constantly [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex] during the fall.
Let [tex]x[/tex] denote the displacement. In this question, [tex]x = (-2.0)\; {\rm m}[/tex]. Note that [tex]x[/tex] is negative since the position after the fall is below the initial position.
The initial velocity at the beginning of the fall is [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] under the assumptions.
Let [tex]v[/tex] denote the final velocity right before landing. Apply the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex] to find [tex]v\![/tex]. Rearrange this equation to obtain:
[tex]\begin{aligned}v &= \sqrt{2\, a\, x + u^{2}} \end{aligned}[/tex].
Substitute in [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex], [tex]x = (-2.0)\; {\rm m}[/tex], and [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] and solve for [tex]v[/tex]
[tex]\begin{aligned}v &= -\sqrt{2\, a\, x + u^{2}} \\ &= -\sqrt{2 \,((-9.81)\; {\rm m\cdot s^{-2}}) \, ((-2.0)\; {\rm m}) + (0\; {\rm m\cdot s^{-1}})^{2}} \\ &= \left(-\sqrt{2 \, (-9.81)\, (-2.0)}\right)\; {\rm m\cdot s^{-1}} \\ &\approx (-6.26)\; {\rm m\cdot s^{-1}} \\ &\approx (-6.3)\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
(Negative since the final velocity points downwards to the ground.)
In other words, the velocity right before landing would be approximately [tex](-6.3)\; {\rm m\cdot s^{-1}}[/tex].
The change in velocity during the fall would be:
[tex]\begin{aligned}v - u &\approx (-6.26)\; {\rm m\cdot s^{-1}} - (0\; {\rm m\cdot s^{-1}}) = (-6.26)\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Divide the change in velocity by acceleration to find the duration [tex]t[/tex] of the fall:
[tex]\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{(-6.26)\; {\rm m\cdot s^{-1}}}{(-9.81)\; {\rm m\cdot s^{-2}}} \\ &\approx 0.64\; {\rm s}\end{aligned}[/tex].
